How To Find The Cardinal Number Of A Set

Primal Number of a set

The number of distinct elements or members in a finite gear up is known every bit the cardinal number of a set. Basically, through cardinality, we define the size of a prepare. The primal number of a set A is denoted as northward(A), where A is any set up and n(A) is the number of members in fix A.

Consider a set A consisting of the prime numbers less than ten.

Set A ={two, 3, v, 7}.

As the set A consists of 4 elements, therefore, the cardinal number of fix A is given every bit north(A) = 4.

Properties related to difference, marriage and intersection and the primal number of set

i) Marriage of Disjoint Sets:

If A and B are two finite sets and if A ∩ B = ∅, then

north(A ∪ B) = n(A) + n(B)

In elementary words if A and B are finite sets and these sets are disjoint then the cardinal number of Union of sets A and B is equal to the sum of the cardinal number of set up A and set B.

Disjoint sets

Figure 1- Disjoint sets

The union of the disjoint sets A and B represented by the Venn diagram is given by A ∪ B and it can be seen that A ∩ B = ∅ because no element is common to both the sets.

ii) Union of 2 sets:

If A and B are two finite sets, then

due north(A ∪ B) = northward(A) + n(B) – n(A ∩ B)

Merely, the number of elements in the union of set A and B is equal to the sum of central numbers of the sets A and B, minus that of their intersection.

Union of two sets

Figure ii- Union of ii sets

In the effigy given above the differently shaded regions depict the different disjoint sets i.e. A – B, B – A and A ∩ B are three disjoint sets as shown and the sum of these represents A ∪ B. Hence,

n (A ∪ B) = n (A – B) + n(B – A) + n(A ∩ B)

iii) Wedlock of 3 sets

If A, B and C are 3 finite sets, and then;

due north(A ∪ B ∪ C) = n(A) + n(B) + north(C) – n(A ∩ B) – n(B ∩ C) – north(A ∩ C) + n(A ∩ B ∩ C)

This is conspicuously visible from the Venn diagram that the union of the three sets will exist the sum of the cardinal number of set A, fix B, fix C and the common elements of the 3 sets excluding the common elements of sets taken in pairs of ii.

Union of three sets

Figure 3-Union of three sets

Video Lesson

Applied Concept – Cardinality of Sets

Solved Case

Let u.s.a. encounter an example to make our point clear.

Example: There is a total of 200 students in course 11. 120 of them study mathematics, 50 students study commerce and 30students study both mathematics and commerce. Notice the number of students who

i) Study mathematics but not commerce

ii) Study commerce but not mathematics

iii) Report mathematics or commerce

Solution: The total number of students represents the cardinal number of the universal set. Let A denote the set of students studying mathematics and fix B represent the students studying commerce.

Therefore,

north (U) = 200

northward(A) = 120

n(B) = fifty

due north(A ∩ B) = 30

The Venn diagram represents the number of students studying mathematics and commerce.
i) Here, nosotros are required to notice the difference of sets A and B.

n(A) = north(A – B) + n(A ∩ B)

n(A-B) = due north(A) – north(A ∩ B)
⇒ north (A – B) = 120 – 30 = 90

The number of students who written report mathematics just not commerce is 90.

ii) Similarly here, we are required to find the divergence of sets B and A

northward (B) = n (B – A) + n (A ∩ B)
⇒ northward (B – A) = 50 – 30 = 20

The number of students who study commerce just not mathematics is 20.

iii) The number of students who study mathematics or commerce

due north (A ∪ B) = n(A) + north(B) – n(A ∩ B)

⇒ northward(A ∪ B) = 120 + 50 – thirty = 140

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